# HackerEarth Distinct Count problem solution

In this HackerEarth Distinct Count problem solution we have given an array A of N integers, classify it as being Good Bad, or Average. It is called Good, if it contains exactly X distinct integers, Bad if it contains less than X distinct integers, and Average if it contains more than X distinct integers.

## HackerEarth Distinct Count problem solution.

`#include <bits/stdc++.h>using namespace std;int main () {    int tc;    scanf("%d",&tc);    while (tc--) {        int n, k;        scanf("%d%d",&n,&k);        set < int > s;        int temp;        for (int i=0; i<n; i++) {            scanf("%d",&temp);            s.insert(temp);        }        if (s.size()<k) {            printf("Bad husband\n");        }        if (s.size()==k) {            printf("Perfect husband\n");        }        if (s.size()>k) {            printf("Lame husband\n");        }    }    return 0;}`

### Second solution

`#include<bits/stdc++.h>using namespace std;#define ll              long long int#define vi              vector<int>#define vl              vector<ll>#define pii             pair<int,int>#define pil             pair<int, ll>#define pll             pair<ll, ll>#define pli             pair<ll, int>#define pb(v, a)        v.push_back(a)#define mp(a, b)        make_pair(a, b)#define MOD             1000000007#define rep(i, a, b)    for(i=a; i<=b; ++i)#define rrep(i, a, b)   for(i=a; i>=b; --i)#define si(a)           scanf("%d", &a)#define sl(a)           scanf("%lld", &a)#define pi(a)           printf("%d", a)#define pl(a)           printf("%lld", a)#define pn              printf("\n")ll pow_mod(ll a, ll b){    ll res = 1;    while(b)    {        if(b & 1)            res = (res * a) % MOD;        a = (a * a) % MOD;        b >>= 1;    }    return res;}set<int> s;int main(){    int t, i, n, x, j, tmp;    si(t);    assert(t >= 1 && t <= 50);    rep(i, 1, t)    {        s.clear();        si(n);        si(x);        assert(n >= 1 && n <= 13000);        assert(x >= 1 && x <= 13000);        rep(j, 1, n)        {            si(tmp);            assert(tmp >= 1 && tmp <= 1000000000);            s.insert(tmp);        }        if(s.size() < x)            printf("Bad husband\n");        else if(s.size() > x)            printf("Lame husband\n");        else            printf("Perfect husband\n");    }    return 0;}`